Solving Systems of Equations by Substitution

Date: 08/24/98 at 19:14:21
From: Lynn
Subject: Solving Systems of Equations by Substitution
Hi. I am currently in the 10th grade at my school and I am having
problems using the substitution method. When they want you to
substitute the x value in place of it in the y problem I totally get
lost. I have even had friends try to help but it doesn't work. I just
don't understand.
Thanks.

Date: 08/25/98 at 16:53:53
From: Doctor Peterson
Subject: Re: Solving Systems of Equations by subsitution
Hi, Lynn.
Substitution can be a little confusing at first, because there are more
things to keep track of than you are used to. But really what you're
doing is keeping things simple by thinking about only one variable at a
time. The idea is that you have two unknown variables (or more). You
solve one equation as if it were a single-variable equation, and feed
its result into the other(s). The trick for beginners is not to be
confused by the idea of solving an equation for one variable while the
other variable is still there.
Let's work through a simple example:
[1] x + y = 1
[2] x - 2y = 3
You can choose either variable to substitute, and either equation to
start with. With enough practice you'll start to see why one order
would be better than another in some cases, but it doesn't really make
a big difference. Let's solve the first equation for y and substitute
it in the second equation.
(I'm numbering the equations to help you follow each equation through
the process. I'll put a ' after a number when I rearrange it.)
I solve the first equation for y:
[1] x + y = 1
[1'] y = 1 - x
That was easy. What does it mean? I'm pretending for the moment that I
know what x is, so I herd it over to the right side along with all the
known values, leaving y alone on the left. After all, if you told me
what x is, I could just plug it into this formula and tell you what y
is. The only problem is that I don't know x - not yet.
Now I substitute "(1 - x)" into the other equation everywhere I see a
"y" since I know now that that is what y is:
[2] x - 2y = 3
[2'] x - 2(1 - x) = 3
Notice how important it is to put those parentheses in. I want to
multiply y by 2, so now I'm multiplying all of 1-x by 2, not just
the 1.
Now I have one equation with only one unknown. From here on it's just
a matter of solving a single equation, and we know how to do that.
Think about what we've just done. We pretended we knew what x was, so
that the first equation could be solved for one variable. Then once
we'd put that value for y into the second equation, we opened our eyes
and realized that we really don't know what x is after all, but now we
know how to find it. Pretty tricky.
Let's finish up. Simplify that new equation, then herd the x over to
the left side by itself:
[2'] x - 2(1 - x) = 3
x - 2 + 2x = 3
3x - 2 = 3
3x = 5
x = 5/3
Now we really do know what x is, so we can go back to that equation
we wrote that claimed to be a solution for y, and put the value of x
in it:
[1'] y = 1 - x
y = 1 - 5/3 = -2/3
That gives us a real value for y.
Now did all this trickery really get us a good solution? Let's check
that both equations really work with x = 5/3 and y = -2/3:
[1] x + y = 5/3 + -2/3 = 3/3 = 1 yup!
[2] x - 2y = 5/3 - 2*(-2/3) = 5/3 + 4/3 = 9/3 = 3 yup!
It all worked. You see, we just worked on one equation at a time, and
with one variable at a time, and we figured it out.
Let me know if that helps. If I did some steps too quickly for you,
maybe part of your problem is that you aren't quite confident enough
in solving a single equation. We can work on that, too, if you show me
where it's hard.
- Doctor Peterson, The Math Forum
Check out our web site! http://mathforum.org/dr.math/